Problem: Simplify the following expression: $y = \dfrac{9x^2- 16x+7}{9x - 7}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(9)}{(7)} &=& 63 \\ {a} + {b} &=& &=& {-16} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $63$ and add them together. The factors that add up to ${-16}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-7}$ and ${b}$ is ${-9}$ $ \begin{eqnarray} {ab} &=& ({-7})({-9}) &=& 63 \\ {a} + {b} &=& {-7} + {-9} &=& -16 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({9}x^2 {-7}x) + ({-9}x +{7}) $ Factor out the common factors: $ x(9x - 7) - 1(9x - 7)$ Now factor out $(9x - 7)$ $ (9x - 7)(x - 1)$ The original expression can therefore be written: $ \dfrac{(9x - 7)(x - 1)}{9x - 7}$ We are dividing by $9x - 7$ , so $9x - 7 \neq 0$ Therefore, $x \neq \frac{7}{9}$ This leaves us with $x - 1; x \neq \frac{7}{9}$.